UC San Diego J. Connelly

ECE 45 Homework 1 Solutions

Problem 1.1 Find the magnitude and phase of the following complex number

(1− j) (2e−jπ/3) (sin(1)) (j2)

(2 + 2j) (−j cos(1)) (ejπ)

without using a calculator. Simplify as much as possible.

Solution:

Recall when taking the product of complex numbers we have XY = |X|ej∠X |Y |ej∠Y , so the magni-tude of the product is |X||Y | and the phase of the product is ∠X + ∠Y . Thus

∣

∣

∣

∣

(1− j) (2e−jπ/3) (sin(1)) (j2)

(2 + 2j) (−j cos(1)) (ejπ)

∣

∣

∣

∣

=|1− j| |2e−jπ/3| | sin(1)| |j2||2 + 2j| | − j cos(1)| |ejπ|

=(√2) (2) (sin(1)) 1

(√8) (cos(1)) 1

=sin(1)

cos(1)= tan(1)

and

∠

(

(1− j) (2e−jπ/3) (cos(1)) (j2)

(2 + 2j) (−j sin(1)) (ejπ)

)

= ∠(1− j) + ∠(2e−jπ/3) + ∠(cos(1)) + ∠(j2)

− ∠(2 + 2j)− ∠(−j sin(1))− ∠(ejπ)

= (−π/4) + (−π/3) + (0) + (π)− (π/4)− (−π/2)− (π)

= −π/3.

Thus the above complex number can be written as:

tan(1)e−jπ/3 =tan(1)

2(1−

√3).

Please report any typos/errors to j2connelly@ucsd.edu

Problem 1.2 Let X = 2 + j + 2e−j2π/3 − ejπ/2.

(a) Find the real portion of X .

(b) Find the phase of the complex conjugate of X .

(c) Let Y = −2 + 2j. Plot X/Y on the complex plane (real and imaginary axes).

Solution:

(a) X = 2 + j + 2 (−1/2− j√3/2)− j = 1− j

√3 ⇒ Real(X) = 1.

(b) X = 1− j√3 ⇒ X∗ = 1 + j

√3 = 2ejπ/3 ⇒ ∠X∗ = π/3.

(c) X = 1− j√3 = 2e−jπ/3 and Y = −2 + 2j = 2

√2ej3π/4. Thus

X

Y=

2 e−jπ/3

2√2 ej3π/4

=

√2

2ej11π/12.

11

Img

Re

22

12

Problem 1.3 Let f(x) =

1 if − 1 ≤ x ≤ 0

−1 if 0 < x < 2

0 otherwise.

(a) Plot f(2x+ 1).

(b) Plot the magnitude of ejx f(x).

(c) If the energy of a signal g(x) is defined to be

E[g(x)] =

∫

∞

−∞

|g(x)|2 dx

how does the energy of g(x) compare to the energy of ejx g(x)?

Solution:

(a) We have

f(2x+ 1) =

1 if − 1 ≤ 2x+ 1 ≤ 0

−1 if 0 < 2x+ 1 < 2

0 otherwise

=

1 if − 1 ≤ x ≤ −1/2

−1 if − 1/2 < x < 1/2

0 otherwise

(b) |ejx| = 1 for all x, so

|ejx f(x)| = |ejx| |f(x)| = |f(x)| =

1 if − 1 ≤ x < 2

0 otherwise

(c) Since |ejx| = 1, for all x, we have

E[g(x)] =

∫

∞

−∞

|g(x)|2 dx =

∫

∞

−∞

|ejx|2|g(x)|2 dx =

∫

∞

−∞

|ejx g(x)|2 dxE[

ejx g(x)]

Problem 1.4 Represent the following sinusoidal functions as phasors.

(a) f1(t) = 3 cos(4t)− 4 sin(4t)

(b) f2(t) = 2(cos(ωt) + cos(ωt+ π/4))

(c) f3(t) = cos2(t)− sin2(t)

Solution:

(a) Taking the phasor transform of f1(t) with frequency 4 yields:

F1 = 3− 4e−jπ/2 = 3 + 4j.

(b) Taking the phasor transform of f2(t) with frequency ω yields:

F2 = 2(1 + ejπ/4).

(c) We need to first verify f4(t) has a single frequency component. Recall

cos2(t) =

(

ejt + e−jt

2

)2

= · · · = 1 + cos(2t)

2

sin2(t) =

(

ejt − e−jt

2j

)2

= · · · = 1− cos(2t)

2

and so f3(t) = cos(2t). Taking the phasor transform of f3(t) with frequency 2 yields F3 = 1.

Problem 1.5 Find the voltage vr(t) in the circuit below, when

(a) vin(t) = 1/3

(b) vin(t) = sin(t)

(c) vin(t) = 1 + 2 sin(t− π).

where R = 1Ω, C = 1/2F , L = 2H .

vin(t) vr(t)R

L

C

Solution:

We need to find the transfer function H(ω) = Vr/Iin.

Assume that iin(t) is sinusoidal with frequency ω. Then by taking the phasor transformation of thecircuit, we have

Vin VRZR

ZL

ZC

where ZR = 2, ZC = 2jω

, and ZL = 2jω.

Voltage divider:

Vr = VinZC//ZR

ZC//ZR + ZL

.

And so

H(ω) =1

1 + ZL/ZR + ZL/ZC=

1

1 + 2jω + (jω)2=

1

(jω + 1)2.

(a) In this case, we have ω = 0 and Vin = 1/3, so

Vr = H(0)Vin = 1/3 −→ vr(t) = 1/3.

(b) In this case, we have ω = 1 and Vin = e−jπ/2, so

Vr = H(1)Vin =e−jπ/2

(√2ejπ4

)2 = −1

2−→ vr(t) = −1

2cos(t).

(c) Here we utilize the fact an RLC circuit is linear (i.e. super-position) and time-invariant.

Let v(1)in (t) = 1/3 and v

(2)in (t) = sin(t) and for each k = 1, 2, let v

(k)r (t) be the voltage across the

resistor when v(k)in (t) is the input voltage.

Then by parts (a) and (b), we have v(1)r (t) = 1/3 and v

(2)r (t) = −1

2cos(t).

We have vin(t) = 3v(1)in (t) + 2v

(2)in (t− π), so by linearity and time invariance, we have

vr(t) = 3v(1)r (t) + 2v(2)r (t− π) = 1− cos(t− π) = 1 + cos(t).

Problem 1.6 Recall the Norton Equivalent of an RLC circuit is a current source in parallel with aresistor and a capacitor or an inductor.

Find the value of C for which the Norton Equivalent is a current source in parallel with only a resistor

(i.e. the Thevenin Impedance is purely real). What is isc(t) and Rth is such a case?

v(t) i(t)

R

LCRth vo(t)vo(t) isc(t)

where v(t) = cos(4t+ π/3), i(t) = sin(4t+ 5π/6), R = 1Ω, and L = 1/4H . C =??

Solution:

V

I

I

Vo

Vo

Vo

ZR

ZR

ZL

ZLZC

ZC

VZR

I + VZR

ZR//ZC//ZL

Since v(t) and i(t) are both sinusoidal

with frequency 4, we can take the phasor

transform with respect to ω = 4.

Then by using a source transformation on

the voltage source and ZR, we have:

Isc = I + V/ZR and

Zeff = ZR//ZC//ZL

=1

1ZR

+ 1ZC

+ 1ZL

=1

1 + j4C − j.

If Zeff is real, then C = 1/4F .

Finally,

isc(t) = i(t) +v(t)

R= 2 cos(4t+ π/3)

Rth = 1Ω.

Alternatively:

We can solve for Isc by shorting the

output terminals:

−+V I

ZR

Isc

We can solve for Zth by setting V = 0and I = 0 and solving for the effective

impedance across the output terminals:

ZR ZL ZC

Problem 1.7 Suppose H(ω) is the transfer function of a linear system and H(ω) = 1 when |ω| < πand H(ω) = 0 otherwise. If the input to the system is

x(t) =

∞∑

k=1

cos(

3π4kt+ π

k

)

k

then what is the output y(t)?

Solution:

For each k = 1, 2, . . . , let xk(t) =cos( 3π

4kt+π

k )k

and let yk(t) denote the output when xk(t) is the input.

Then x(t) =∑

∞

k=1 xk(t), so by the linearity of the system, y(t) =∑

∞

k=1 yk(t).

For each k = 1, 2, . . . , since xk(t) is sinusoidal with frequency 3π4k, we have

yk(t) =

∣

∣H(

3π4k)∣

∣

kcos

(

3π

4kt +

π

k+ ∠H

(

3π

4k

))

=

cos(3π4t+ π) k = 1

0 else

Thus y(t) = y1(t) = − cos(3π4t).

Problem 1.8 Are the following steady-state input-output pairs consistent with the properties of RLCcircuits (or more generally, LTI systems)?

(a) cos(2t) → H(ω) → 99 sin(2t− e)

(b) cos(4t) → H(ω) → 1 + 4 cos(4t)

(c) 4 → H(ω) → −8

(d) 4 → H(ω) → 8j

(e) 4 → H(ω) → cos(3t)

(f) sin(πt) → H(ω) → cos(πt) + sin(πt)

(g) sin(πt) → H(ω) → sin2(πt)

(h) 0 → H(ω) → 5.

Solution:

(a) Yes. The output is an amplitude-scaled and phase-shifted version of the input.

(b) No. There is a new frequency component, ω = 0 in the output.

(c) Yes. The output is an amplitude-scaled and phase-shifted version of the input.

(d) No. We cannot have imaginary values as output signals when the input is real.

(e) No. There is a new frequency component, ω = 3 in the output.

(f) Yes. The output is an amplitude-scaled and phase-shifted version of the input,

since cos(πt) + sin(πt) =√2 cos(πt− π/4) .

(g) No. There are two new frequency components, ω = 0 and ω = 2π,

since sin2(πt) = 12(1− cos(2πt)).

(h) No. We cannot have a non-zero output when the input is 0.

Problem 1.9 For each k = 1, 2, suppose yk(t) is the output when xk(t) is the input to an LTI system.

y1(t) =

1 if 0 ≤ t < 20 otherwise

y2(t) =

1 if 0 ≤ t < 1−1 if 1 ≤ t < 20 otherwise.

Determine a possible input to the LTI system (in terms of x1(t) and x2(t)) that could have yielded thefollowing outputs:

(a) z1(t) =

−1 if 2 ≤ t < 40 otherwise

(b) z2(t) =

1 if 1 ≤ t < 50 otherwise

(c) z3(t) =

1 if 0 ≤ t < 10 otherwise

(d) z4(t) =

1 if 3 ≤ t < 40 otherwise

(e) z5(t) =

1 if t ≥ 00 otherwise

(f) z6(t) = 0.

Solution:

(a) z1(t) = −y1(t− 2), so a possible input was −x1(t− 2).

(b) z2(t) = y1(t− 1) + y1(t− 3), so a possible input was x1(t− 1) + x1(t− 3).

(c) z3(t) =12(y1(t) + y2(t)), so a possible input was 1

2(x1(t) + x2(t)).

(d) z4(t) = z3(t− 3)12(y1(t− 3) + y2(t− 3)), so a possible input was 1

2(x1(t− 3) + x2(t− 3)).

(e) z5(t) = y1(t) + y1(t− 2) + y1(t− 4) + · · · , so a possible input was∑

∞

k=0 xk(t− 2k).

(f) z6(t) = 0 = y1(t)− y1(t), so a possible input was x1(t)− x1(t) = 0.

Remark:

For an arbitrary LTI system, suppose y(t) is the output when x(t) is the input. Then when 0 = x(t) −x(t) is the input to the LTI system, by linearity, the output is y(t)− y(t) = 0.

Thus for any LTI system, if the input is 0, then the output must be 0.

However, the converse need not be true (i.e. if the output is 0, then the input is not necessarily 0). Whatare some simple examples of this? (e.g. High-pass filter with a DC input).

Problem 1.10 An LTI system with input x(t) and output y(t) is given by the differential equation

3d4y(t)

dt4− 2

d3x(t)

dt+ y(t) = 2x(t) +

d2y(t)

dt.

Find the steady-state output when x(t) = 1 + cos(t) + cos(2t).

Solution:

Let x0(t) = 1, x1(t) = cos(t), and x2(t) = cos(2t),

For each k = 0, 1, 2, let yk(t) be the output when xk(t) is the input.

Then x(t) = x0(t) + x1(t) + x2(t), so by the linearity of the system, y(t) = y0(t) + y1(t) + y2(t).

We can find the transfer function of the system by taking the phasor transformation of both sides of thedifferential equation:

3(jω)4Y − 2(jω)3X + Y = 2X + 2(jω)2Y

⇒ H(ω) =Y

X=

2− 2jω3

3ω4 + 2ω2 + 1

For each k = 0, 1, 2, let Xk and Yk be the phasor transformations of xk(t) and yk(t) with frequency k,respectively. Then

Y0 = H(0)X0 = 2

⇒ y0(t) = 2

Y1 = H(1)X1 =

√2

3e−jπ/4

⇒ y1(t) =

√2

3cos(t− π/4)

Y2 = H(2)X2 =2− 16j

57=

√260

57e−j tan−1(8)

⇒ y2(t) =

√260

57cos(2t− tan−1(8)).

Finally, y(t) = y0(t) + y1(t) + y2(t) = 2 +

√2

3cos(t− π/4) +

√260

57cos(2t− tan−1(8)).

Problem 1.11 Recall a low-pass filter only allows frequencies below some threshold, a high-pass

filter only allows frequencies above some threshold, a band-pass filter only allows frequencies in somerange, and a band-reject filter only allows frequencies outside of some range.

What type of filter are the following circuits? Justify your answer by finding the magnitude of the

transfer function of each circuit.

(a) (b)

RR CC io(t)io(t) iin(t)iin(t)

Solution:

(a) We can use a current divider to get:

Io = Iin1/ZR

1/ZR + 1/ZC

⇒ H(ω) =IoIin

=1

1 + ZR/ZC=

1

1 + jωRC

⇒ |H(ω)| = 1√

1 + (ωRC)2

When ω ≈ 0, |H(ω)| ≈ 1, and as ω → ∞, |H(ω)| ≈ 0. Thus this is a low-pass filter.

(b) We can again use a current divider to get:

Io = Iin1/ZC

1/ZR + 1/ZC

⇒ H(ω) =IoIin

=1

1 + ZC/ZR

=1

1 + 1jωRC

⇒ |H(ω)| = 1√

1 +(

1ωRC

)2

When ω ≈ 0, |H(ω)| ≈ 0, and as ω → ∞, |H(ω)| ≈ 1. Thus this is a high-pass filter.

Problem 1.12 For each of the circuits in the previous problem, let RC = 100.

If the circuit is a low-pass filter, find the frequency ωc such that |H(ω)| < 0.01 for all ω > ωc.

If the circuit is a high-pass filter, find the frequency ωc such that |H(ω)| < 0.01 for all ω < ωc.

If RC = β for some β > 0, what is ωc in terms of β?

Solution:

(a) The magnitude of the low-pass filter is given by

|H(ω)| = 1√

1 + (ωRC)2

Suppose |H(ω)| < 1100

. Then we have

√

1 + (ωRC)2 > 100

⇒ ω2 >1002 − 1

(RC)2≈

(

100

RC

)2

⇒ ω >100

β.

(a) The magnitude of the high-pass filter is given by

|H(ω)| = 1√

1 +(

1ωRC

)2

Suppose |H(ω)| < 1100

. Then we have

√

1 +

(

1

ωRC

)2

> 100

⇒ ω2 <1

(RC)2(1002 − 1)≈

(

1

100RC

)2

⇒ ω <1

100β.

MATLAB Problem 1 Download and load the file “sum.mat” into MATLAB by placing it in yourMATLAB directory and running “load sum;” The entries of the array sum are given as follows:

sum[1] = z[1]

sum[2] = z[1] + z[2]

sum[3] = z[1] + z[2] + z[3]

...

sum[N − 1] = z[1] + z[2] + · · ·+ z[N − 1]

sum[N ] = z[1] + z[2] + · · ·+ z[N ]

where z is an array containing an audio message. In other words, the kth entry of sum is the sum ofthe first k entries of z. Your goal is to use MATLAB to recover the audio message z from the arraysum and play it within MATLAB, using “sound(z, Fs);” where Fs = 11025. The array z has the same

length as the array sum.

• What is the (well-known) audio message? Include the code you used to decipher z.

• In MATLAB, let N = length(sum); and define the array t = (1 : N)/Fs; Include the output of:“plot(t, z);” and label:

– the x-axis as “Time (seconds)”

– the y-axis as “Amplitude”

– the title as whatever you deem appropriate.

• When running “sound(z, Fs);” what happens if you set Fs = 5000, instead of 11025? What ifFs = 20000?

Solutions

Since sum[n] =n

∑

k=1

z[n], we can recover z[n] as follows:

z[1] = sum[1] and z[n] = sum[n]− sum[n− 1] for n ≥ 2.

See ECE45 MATLAB1 1.m for MATLAB script.

“If only you knew the power of the Dark Side.”- Darth Vader (Star Wars: Episode V - The Empire Strikes Back )

When Fs = 5000, MATLAB is assuming the signal was sampled at 5000 Hz (when in fact, it wassampled at 11025 Hz), so its playback speed is roughly halved. Similarly, when Fs = 20000, theplayback speed is roughly doubled.

Fs is the frequency used to sample to .wav file I used to generate this signal. When you play back an audio file in

MATLAB, it needs to know the frequency at which the signal was sampled. We will talk more about sampling later in the

course.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

"If you only knew the power of the Dark Side"

Time (seconds)

Am

plitu

de

MATLAB Problem 2 Let N be a positive integer, and define the length-4000 arrays x1, x2, and x3 asfollows:

x1[k] =1

2− 1

π

N∑

n=1

sin(

πn500

k)

n

x2[k] =2

π− 4

π

N∑

n=1

cos(

πn500

k)

4n2 − 1

x3[k] =1

2+

2

π

2N−1∑

n=1n odd

sin(

πn500

k)

n

where xi[k] refers to the kth entry of the array xi and k ranges from 1 to 4000.

• Create an array t = (1 : 4000)/1000; and include the output of “plot(t, x1);” “plot(t, x2);” and“plot(t, x3);” when N = 1, 5, 10, and 100. Be sure to label your plots.

• What do you notice about these functions as N grows large? In each case, what function is beingapproximated?

Solutions

0 1 2 3 4

0

0.5

1

x1(t), N = 1

0 1 2 3 4

0

0.5

1

x1(t), N = 5

0 1 2 3 4

0

0.5

1

x1(t), N = 10

0 1 2 3 4

0

0.5

1

x1(t), N = 100

As N grows large, x1(t) converges to a Sawtooth wave that repeats every 1 time unit.

0 1 2 3 4

0

0.5

1

x2(t), N = 1

0 1 2 3 4

0

0.5

1

x2(t), N = 5

0 1 2 3 4

0

0.5

1

x2(t), N = 10

0 1 2 3 4

0

0.5

1

x2(t), N = 100

As N grows large, x2(t) converges to | sin(2πt)|.

0 1 2 3 4

0

0.5

1

x3(t), N = 1

0 1 2 3 4

0

0.5

1

x3(t), N = 5

0 1 2 3 4

0

0.5

1

x3(t), N = 10

0 1 2 3 4

0

0.5

1

x3(t), N = 100

As N grows large, x3(t) converges to a square wave that repeats every 1 time unit.

In each case, as N grows, we are adding sinusoids with higher and higher frequencies to the signalxi(t), which yields a better approximation of the desired signal.